MATH529: L13 Integral transforms

Overview

Integral transforms provide solutions to linear PDEs, covering cases in which separation of variables fails

• Error function

• Laplace transform

Error function

• Error function, $\mathrm{erf}\left(x\right)$, complementary error function $\mathrm{erfc}\left(x\right)$

  ${\displaystyle \mathrm{erf}\left(x\right)=\frac{2}{\sqrt{\pi }}\underset{0}{\overset{x}{\int }}{e}^{-u^2}du,\mathrm{erfc}\left(x\right)=\frac{2}{\sqrt{\pi }}\underset{x}{\overset{\infty }{\int }}{e}^{-u^2}du}$ ${\displaystyle {lim}_{x\to \infty }\mathrm{erf}\left(x\right)=1,\mathrm{erf}\left(x\right)+\mathrm{erfc}\left(x\right)=1}$

Laplace transform

• Used for time-unsteady problems, initial value problems

• $f:ℝ\to ℝ$, the Laplace transform is

  ${\displaystyle F\left(s\right)=ℒ\left\{f\left(t\right)\right\}=\underset{0}{\overset{\infty }{\int }}{e}^{-st}f\left(t\right)dt}$

  and can be considered as the decaying-function analog to a Fourier series

  ${\displaystyle f:ℝ\to ℝ,f(t+2\pi )=f\left(t\right),f\left(t\right)=\frac{1}{2}{A}_0+\sum _{k=1}^{\infty }[A_k\cos (kt)+B_k\sin (kt)]}$

• Let $F\left(s\right)=ℒ\left\{f\left(t\right)\right\}$. Properties

  ${\displaystyle ℒ\left\{f^{\text{'}}\left(t\right)\right\}=\underset{0}{\overset{\infty }{\int }}{e}^{-st}{f}^{\text{'}}\left(t\right)dt=-f\left(0\right)+sF\left(s\right)}$

Laplace transform properties

• Repeated differentiation $F\left(s\right)=ℒ\left\{f\left(t\right)\right\}$

  ${\displaystyle ℒ\left\{f^{{}^{\text{'}\text{'}}}\left(t\right)\right\}=sℒ\left\{f^{{}^{\text{'}}}\left(t\right)\right\}-f^{\text{'}}\left(0\right)=s^{2}F\left(s\right)-sf\left(0\right)-f^{\text{'}}\left(0\right)}$ ${\displaystyle ℒ\left\{f^{\left(k\right)}\left(t\right)\right\}=s^{k}F\left(s\right)-s^{k-1}f\left(0\right)-s^{k-2}{f}^{\text{'}}\left(0\right)-\cdots -f^{(k-1)}\left(0\right)}$

• Apply to a PDE, $u_{tt}=a^2{u}_{xx}$, $u(x,t)$, $U(x,s)=$${\int }_0^{\infty }{e}^{-st}u(x,t)dt$

  ${\displaystyle ℒ\left\{u_{tt}\right\}=ℒ\{a^2{u}_{xx}\}\iff \underset{0}{\overset{\infty }{\int }}{e}^{-st}{u}_{tt}(x,t)dt=a^2\underset{0}{\overset{\infty }{\int }}{e}^{-st}{u}_{xx}(x,t)dt}$ ${\displaystyle s^{2}U(x,s)-su(x,0)-\frac{\partial u}{\partial t}(x,0)=a^2\frac{{\partial }^2}{\partial x^2}U(x,s)}$